Math 100 Ÿ105, Fall Term 2010
Midterm Exam
November 8th ,2010
Student number:
LAST name:
First name:
Instructions
• Do not turn this page over until instructed. You will have 45 minutes for the exam.
• You may not use books, notes or electronic devices of any kind.
• Solutions should be written clearly, in complete English sentences, showing all your work.
• If you are using a result from the textbook, the lectures or the problem sets, state it properly.
Signature:
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Total
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/18
/6
/6
/10
/40
Math 100 Ÿ105, Fall 2010
1
Midterm, page 2/7
Problem 1
Short-form answers
Show your work and clearly delineate your nal answer. Not all problems are of equal diculty.
[3] a. Dierentiate the function y = arctan(x); write your answer as a function of
u
alone (you may use the formula d tan
= 1 + tan2 u).
du
x
If y = arctan(x) then tan(y) = x. Dierentiating with respect to x and using the chain rule we nd
y)
dy
dy
dy
1 = d(tan
= (1 + tan2 y) dx
= (1 + x2 ) dx
. Solving for dx
we nd
dx
1
dy
.
=
dx
1 + x2
[3] b. Find the tangent line to y = xsin x at the point (π, 1).
Taking logarithms we have ln y = sin x ln x, and dierentiating we nd
1 dy
sin x
= cos x ln x +
.
y dx
x
When x = π we have y = 1 and sin π = 0 so
y 0 (π) = cos π ln π = − ln π .
The tangent line is therefore
y = 1 − (x − π) ln π .
[3] c. We have z(t) = ex(t)·y(t) where x, y both depend on t. If at t = 1 we have x(1) = 0,
x0 (1) = 1, y(1) = 2, y 0 (1) = 3 nd dz
dt at t = 1.
Dierentiating using the chain rule we have:
dz
d
= ex(t)y(t) (xy)
dt
dt
= exy (x0 y + y 0 x) .
At t = 1 we have x(1) = 0 so
z 0 (1) = e0 x0 (1)y(1) = 1 · 1 · 2 = 2 .
Math 100 Ÿ105, Fall 2010
Midterm, page 3/7
Problem 1
[3] d. The population of Canada was roughly 18 million in the year 1960, 30 million in
the year 2000. Assuming the population grows exponentially, estimate the population
of Canada in 2040.
5
In 40 years the population grew by a factor of about 30
18 = 3 . We thus expect the population in
2040 to be about
5
· 30million = 50million .
3
[3] e. Let f (x) = ex + e−x . Use a 2nd order Taylor polynomial to give a rational number
approximating f ( 12 ).
To second order we have ex ≈ 1 + x +
we nd
At x =
1
2
x2
2
so e−x ≈ 1 + (−x) +
(−x)2
2
= 1−x+
x2
2 .
Adding the two
ex + e−x ≈ 2 + x2 .
this reads
1
e1/2 + e−1/2 ≈ 2 .
4
[3] f. Show that the error in the approximation is less than
that 2 ≤ e ≤ 3.
1
50
. You may use the fact
We note that the expansion above is correct to 3rd order (the next term would be
error is therefore at most
(4) 4
f (c) 1
24
2
x3
6
3
+ (−x)
6 ). The
for some 0 ≤ c ≤ 12 . Now the fourth derivative of ex is ex , and of e−x is (−1)4 e−x = e−x so
√
√
f (4) (x) = f (x) = ex + e−x . For 0 ≤ c ≤ 21 we then have f (4) (c) = ec + e−c ≤ e + e0 ≤ 4 + 1 = 3.
It follows that the error is at most
1
1
1
3 1
·
=
=
<
.
24 16
8 · 16
128
50
Math 100 Ÿ105, Fall 2010
2
Midterm, page 4/7
Problem 2
Long-form answers
[6] Find the maximum value of f (x) = x
q
1 − 34 x2
on the interval [0, 1].
q
• For |x| < 1 we have 43 x2 ≤ 34 < 1 so by the chain rule 1 − 34 x2 is dierentiable in the open
interval (0, 1). By the product rule the same holds for f . f is also continuous where dened,
so it is continuous in the closed interval. It follow that the maximum of f occurs either at an
endpoint or at a critical point.
q
• We have f (0) = 0 and f (1) = 1 − 43 = 12 .
• We have f 0 (x) =
q
1 − 43 x2 + x √
− 32 x =
q
f 0 (x0 ) = 0 when 32 x20 = 1, that is for x0 = 23 .
2
1
1− 34 x2
q q
q
• At the critical point, we have f ( 23 ) = 23 1 −
• Finally, since 3 < 4 we have
√
3<
√
4 so
√1
3
>
1
2
3
4
·
1− 34 x2 − 43 x2
√
2
3
=
1− 43 x2
q
2
3
3
2
1− x
= √ 2 3 2 .It follows that
1−
1− 4 x
1
2
=
√1 .
3
> 0 so the absolute maximum is
√1 .
3
Math 100 Ÿ105, Fall 2010
3
Midterm, page 5/7
Problem 3
Long-form answers
[6] A point is moving on the curve y2 = x3 − 3 in such a way that the x-co-ordinate
. √How fast is the distance of the point to the origin
is changing at the rate of 3
changing, when the point is at (2, 5)?
units
min
The distance of the point from the origin is
p
p
D(t) = x2 + y 2 = x2 + x3 − 3 .
It follows that
dD
1
d 2
= √
(x + x3 − 3)
2
3
dt
dt
2 x +x −3
dx
2x + 3x2
= √
.
2
3
2 x + x − 3 dt
When x = 2 and
dx
dt
= 3 units
this gives
min
dD
4 + 12
units
= √
·3
dt
min
2 4+8−3
units
16
= √ ·3
min
2 9
units
.
=8
min
Math 100 Ÿ105, Fall 2010
4
Midterm, page 6/7
Problem 4
Long-form answers
Let f (x) = ex − e−x .
[1] Verify that (f 0 (x))2 = 4 + (f (x))2 .
2
We have f 0 (x) = ex + e−x . Since ex · e−x = ex−x = 1 we have (f (x)) = e2x − 2 + e−2x and
2
(f 0 (x)) = e2x + 2 + e−2x = 4 + e2x − 2 + e−2x .
[3] Show that f has an inverse function.
Three approaches:
2
1. From the rst part we know that (f 0 (x)) ≥ 4 and in particular it is never zero. Since
f 0 (0) = 1 + 1 = 2 > 0 we nd that f 0 (0) > 0 everywhere (f 0 is continuous, so if it took
negative and positive values it would vanish). It follows that f is strictly monotone and so
has an inverse function.
2. We know that ex > 0 for all x. It follows that f 0 (x) = ex + e−x > 0 everywhere.
3. Let y be any real number. We will show that that is a unique x such that f (x) = y . Consider
the equation t − 1t = y for an unkown t > 0 (t stands for ex ), equivalently t2 − yt − 1 = 0.
√
q q y± y 2 +4
y
y 2
y 2
+
1
.
Since
+ 1 > y2 the solution
This has the two solutions
=
±
2
2
2
2
q q y
y 2
y 2
y
−
+
1
is
negative.
On
the
other
hand,
the
solution
t
=
+
+ 1 is positive
2
2
2
2
1
(even if y is negative). So there is a unique positive value of t such that t + t = y . Given t,
there is a unique x such that ex = t (that is x = ln t) so indeed for every y there is a unique
x such that f (x) = y in other words, f has an inverse function. That function is
"
#
r y 2
y
x = ln
+
+1 .
2
2
Math 100 Ÿ105, Fall 2010
[3] Let
x, y .
x = g(y)
Midterm, page 7/7
Problem 4
be the inverse function. Find a formula for its derivative in terms of
By the inverse function rule we have
g 0 (y) =
1
f 0 (x)
=
ex
1
.
+ e−x
If you took the third approach before and found g(y) = ln
g (y) =
y
2
=
y
2
y
2
+
q
y 2
2
"
1
=p
.
y2 + 4
[3] Find a formula for g0 (y) involving y alone.
From the rst part we know that f (x) =
0
+ 1 , you would now nd
#
y/2
1
q + p
2 2 (y/2)2 + 1
y 2
+
1
+
2
"p
#
(y/2)2 + 1 + y/2
1
q p
y 2
2
(y/2)2 + 1
+1
+
2
1
0
q
2
4 + (f (x)) =
g 0 (y) = p
1
4 + y2
.
p
4 + y 2 so